If y=a cos (log x)+b sin (log x),where a, b a | vedclass

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(B) Given,$y = a \cos(\log x) + b \sin(\log x)$.First,differentiate with respect to $x$:$y^{\prime} = -a \sin(\log x) \cdot \frac{1}{x} + b \cos(\log

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$-y$

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(B) Given,$y = a \cos(\log x) + b \sin(\log x)$.First,differentiate with respect to $x$:$y^{\prime} = -a \sin(\log x) \cdot \frac{1}{x} + b \cos(\log x) \cdot \frac{1}{x} = \frac{-a \sin(\log x) + b \cos(\log x)}{x}$.This implies $x y^{\prime} = -a \sin(\log x) + b \cos(\log x)$.Now,differentiate again with respect to $x$ using the product rule on the left side:$x y^{\prime \prime} + y^{\prime} = -a \cos(\log x) \cdot \frac{1}{x} - b \sin(\log x) \cdot \frac{1}{x}$.Multiply the entire equation by $x$:$x^2 y^{\prime \prime} + x y^{\prime} = -a \cos(\log x) - b \sin(\log x)$.Factor out the negative sign:$x^2 y^{\prime \prime} + x y^{\prime} = -[a \cos(\log x) + b \sin(\log x)]$.Since $y = a \cos(\log x) + b \sin(\log x)$,we get:$x^2 y^{\prime \prime} + x y^{\prime} = -y$.

If $y=a \cos (\log x)+b \sin (\log x)$,where $a, b$ are parameters,then $x^2 y^{\prime \prime}+x y^{\prime}$ is equal to

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